Word Ladder II

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:

  1. Only one letter can be changed at a time
  2. Each intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

Return

  [
    ["hit","hot","dot","dog","cog"],
    ["hit","hot","lot","log","cog"]
  ]

Note:
  • All words have the same length.
  • All words contain only lowercase alphabetic characters. 
     1 class Node {
 2     int no;
 3     String val;
 4     LinkedList<Node> prev;    
 5     
 6     Node(int no, String v) {
 7         this.no = no;
 8         this.val = v;
 9     }
10     
11     void addPrev(Node pNode) {
12         if (prev == null) {
13             prev = new LinkedList<Node>();
14         }
15         prev.add(pNode);
16     }
17 }
18 
19 public class Solution {    
20     ArrayList<ArrayList<String>> answer;
21     
22     public void findPath(Node node, ArrayList<String> cur, String start) {
23         if (node.val.equals(start)) {
24             answer.add(cur);
25             return;
26         }
27         ArrayList<String> temp;
28         for (Node n : node.prev) {
29             temp = new ArrayList<String>(cur);
30             temp.add(0, n.val);
31             findPath(n, temp, start);
32         }
33     }
34     
35     public ArrayList<ArrayList<String>> findLadders(String start, String end, HashSet<String> dict) {
36         // Start typing your Java solution below
37         // DO NOT write main() function
38         HashMap<String, Node> map = new HashMap<String, Node>();
39         Queue<Node> queue = new LinkedList<Node>();
40         Node node = new Node(0, start);
41         Node endNode = null;
42         map.put(start, node);
43         queue.add(node);
44         boolean stop = false;
45         while (queue.size() > 0 && !stop) {
46             int count = queue.size();
47             for (int i = 0; i < count; i++) {
48                 node = queue.poll();
49                 for (int j = 0; j < node.val.length(); j++) {
50                     StringBuilder t = new StringBuilder(node.val);
51                     for (char k = 'a'; k <= 'z'; k++) {
52                         t.setCharAt(j, k);
53                         if (dict.contains(t.toString())) {
54                             Node v = map.get(t.toString());
55                             if (v == null) {
56                                 Node temp = new Node(node.no + 1, t.toString());
57                                 temp.addPrev(node);
58                                 queue.add(temp);
59                                 map.put(t.toString(), temp);
60                                 if (t.toString().equals(end)) {
61                                     endNode = temp;
62                                     stop = true;
63                                 }
64                             }
65                             else {
66                                 if (v.no == node.no + 1) {
67                                     v.addPrev(node);
68                                 }
69                             }
70                         }
71                     }
72                 }
73             }
74         }
75         answer = new ArrayList<ArrayList<String>>();
76         if (endNode != null) {
77             findPath(endNode, new ArrayList<String>(Arrays.asList(end)), start);
78         }
79         return answer;
80     }
81 }
    View Code

     

posted @ 2014-02-24 05:01  krunning  阅读(226)  评论(0)  编辑  收藏  举报